I am trying to see, if the forgetful functor from $\mathbb{Z}[X]$-modules to abelian groups is exact and in case it is not exact, is it left or right exact.
In general, i understand the definition of a functor. In my example the forgetful functor "forgets" the scalar multiplication in $\mathbb{Z}[X]$, but remembers the addition.
My problems is that i don't know how to apply these definitions to a concrete example.
I tried here with something: ($G$ and $G'$ are abelian groups)
If the forgetful functor is left exact, then we start the sequence with the trivial kernel $0\rightarrow \mathbb{Z}[X]\overset{f}{\rightarrow}G\overset{g}{\rightarrow}G'$, here i have to check that $f$ is injective and $im(f)=ker(g)$.
If it was right exact, we should have $\mathbb{Z}[X]\overset{f}{\rightarrow}G\overset{g}{\rightarrow}G'\rightarrow 0$, where $g$ must be surjective and $im(f)=ker(g)$.
How can we show this formal? Can somene help me with this example, please?
Thank you in advance!
Best Answer
If $R$ is any ring, then the forgetful functor $\mathsf{Mod}(R) \to \mathsf{Ab}$ is exact. Remember that exact := left exact + right exact := preservation of finite limits and finite colimits. In fact, the forgetful functor preserves all limits and colimits. The reason is that one really constructs limits and colimits of $R$-modules as the limit and colimits of the underlying abelian groups and endows them with an $R$-module structure. Here is a more concise explanation: Left adjoints preserve colimits, and right adjoints preserve limits. The forgetful functor has a right adjoint mapping an abelian group $A$ to the $R$-module $\hom_{\mathbb{Z}}(R,A)$. It has a left adjoint mapping an abelian group $A$ to the $R$-module $R \otimes_{\mathbb{Z}} A$.
Edit: If you use the definition of exact functors from homological algebra (not general category), i.e. preservation of short exact sequences, notice that a sequence of $R$-modules $A \to B \to C$ is exact iff the image of $A \to B$ is the kernel of $B \to C$, but that these images and kernels are just the ones for the underlying homomorphisms of abelian groups. Thus, $\mathsf{Mod}(R) \to \mathsf{Ab}$ preserves exact sequences.